Variance-Based Subgradient Extragradient Method for Stochastic Variational Inequality Problems

Abstract

In this paper, we propose a variance-based subgradient extragradient algorithm with line search for stochastic variational inequality problems by aiming at robustness with respect to an unknown Lipschitz constant. This algorithm may be regarded as an integration of a subgradient extragradient algorithm for deterministic variational inequality problems and a stochastic approximation method for expected values. At each iteration, different from the conventional variance-based extragradient algorithms to take projection onto the feasible set twicely, our algorithm conducts a subgradient projection which can be calculated explicitly. Since our algorithm requires only one projection at each iteration, the computation load may be reduced. We discuss the asymptotic convergence, the sublinear convergence rate in terms of the mean natural residual function, and the optimal oracle complexity for the proposed algorithm. Furthermore, we establish the linear convergence rate with finite computational budget under both the strongly Minty variational inequality and the error bound condition. Preliminary numerical experiments indicate that the proposed algorithm is competitive with some existing methods.

Appendices

Appendix A: Proof of Lemma 4.3

First of all, we use Lemma 2.6 with $p=q$, $x^k\in \mathcal{F}_k$ and noting that $\xi ^k$ is independent of $\mathcal{F}_k$, we have

$$\begin{aligned} \big |\Vert \widehat{\epsilon }_1^k\Vert \big |\mathcal{F}_k\big |_p\le C_p\frac{\sigma _{p}(x^*)+L_p\Vert x^k-x^*\Vert }{\sqrt{N_k}}. \end{aligned}$$

(A.1)

Applying Lemma 2.6 again with $p=q$, $y^k\in \mathcal{F}_k$ and noting that $\eta ^k$ is independent of $\mathcal{F}_k$ and $\Big |\big |\cdot |\widehat{\mathcal{F}}_k\big |_p|\mathcal{F}_k\Big |_p=|\cdot |\mathcal{F}_k|$, we have

$$\begin{aligned} \big |\Vert \widehat{\epsilon }_2^k\Vert \big |\mathcal{F}_k\big |_p=\Big |\big |\Vert \widehat{\epsilon }_2^k\Vert \big |\widehat{\mathcal{F}}_k\big |_p\big |\mathcal{F}_k\Big |_p\le C_p\frac{\sigma _{p}(x^*)+L_p\big |\Vert y^k-x^*\Vert \big |\mathcal{F}_k\big |_p}{\sqrt{N_k}}. \end{aligned}$$

(A.2)

From Lemma 2.7, Assumption 4.2 and noting that $0<\alpha _k\le \widehat{\alpha }\le 1$, $y^k=y(x^k,\alpha _k,\xi ^k)$, $x^k\in \mathcal{F}_k$, $\xi ^k$ is independent of $\mathcal{F}_k$, we have

$$\begin{aligned} \big |\Vert \widehat{\epsilon }_3^k\Vert \big |\mathcal{F}_k\big |_p\le \frac{c_1\sigma _{2p}(x^*)+\widehat{L}_{2p}\Vert x^k-x^*\Vert }{\sqrt{N_k}}.\end{aligned}$$

(A.3)

On the other hand, putting together Step 3 in Algorithm 1, Lemmas 2.1 and 2.3, we obtain

$$\begin{aligned} \Vert x^*-y^k\Vert= & {} \Vert \Pi _X[x^*-\alpha _kF(x^*)]-\Pi _X[x^k-\alpha _k\widehat{F}(x^k,\xi ^k)]\Vert \\\le & {} \Vert x^*-\alpha _kF(x^*)-x^k+\alpha _k\widehat{F}(x^k,\xi ^k)\Vert \\\le & {} (1+\widehat{\alpha } L)\Vert x^k-x^*\Vert +\widehat{\alpha }\Vert \widehat{\epsilon }_1^k\Vert . \end{aligned}$$

Taking $|\cdot |\mathcal{F}_k|_p$ in the above inequality, we have

$$\begin{aligned} \big |\Vert x^*-y^k\Vert \big |\mathcal{F}_k\big |_p\le (1+\widehat{\alpha } L)\Vert x^k-x^*\Vert +\widehat{\alpha }\big |\Vert \widehat{\epsilon }_1^k\Vert \big |\mathcal{F}_k\big |_p.\end{aligned}$$

(A.4)

Putting together relations (A.1)–(A.4) and use the facts that $\big |a^2\big |\mathcal{F}_k\big |_{\frac{p}{2}}=\big |a\big |\mathcal{F}_k\big |_p^2,\ (a+b)^2\le 2a^2+2b^2,\ \widehat{L}_{2p}>L_pC_p,\ c_1>C_p$ and $\sigma _{2p}(x^*)\ge \sigma _p(x^*)$, the claim with $\mathcal{C}_p:=2c_1^2\left( 7-3\lambda ^2+\sup _k\frac{12L_p^2\widehat{\alpha }^2C_p^2}{N_k}\right)$ and $\overline{\mathcal{C}}_p:=2\left( 4-3\lambda ^2+6(1+\widehat{\alpha } L)^2+\sup _k\frac{12L_p^2\widehat{\alpha }^2C_p^2}{N_k}\right)$ is proved. $\square$

1.1 Appendix B: Proof of Lemma 4.4

By $y^k\in \widehat{\mathcal{F}}_k$ and $\eta ^k$ is independent of $\widehat{\mathcal{F}}_k$, we have

$$\begin{aligned}\mathbb {E}[\widehat{\epsilon }^k_2\big |\widehat{\mathcal{F}}_k]= & {} \mathbb {E}\Big [\frac{1}{N_k}\sum _{j=1}^{N_k}f(\eta _j^k,y^k)-F(y^k)\Big |\widehat{\mathcal{F}}_k\Big ]\\= & {} \frac{1}{N_k}\sum _{j=1}^{N_k}\mathbb {E}\big [f(\eta _j^k,y^k)\big |\widehat{\mathcal{F}}_k\big ]-F(y^k)\\= & {} \frac{1}{N_k}\sum _{j=1}^{N_k}F(y^k)-F(y^k)=0. \end{aligned}$$

Taking $\mathbb {E}[\cdot |\mathcal {F}_k]$ in (4.1), we can get the conclusion from Lemmas 4.1 and 4.3 immediately.

Appendix C: Proof of Theorem 4.1

By Lemma 4.4, we have

$$\begin{aligned} \mathbb {E}[\Vert x^{k+1}-x^*\Vert ^2\big |\mathcal{F}_k]\le \left( 1+\frac{\overline{\mathcal{C}}_2\widehat{\alpha }^2\widehat{L}^2_4}{N_k}\right) \Vert x^k-x^*\Vert ^2-\rho \mathfrak {R}^2(F,x^k)+\frac{\mathcal{C}_2\widehat{\alpha }^2\sigma _4^2(x^*)}{N_k}. \end{aligned}$$

Taking into account $\sum _kN_k^{-1}<\infty$. By applying Lemma 2.4 with $v^k:=\Vert x^k-x^*\Vert ^2,\ a_k:=\frac{\overline{\mathcal{C}}_2\widehat{\alpha }^2\widehat{L}^2_4}{N_k},\ b_k:=\frac{\mathcal{C}_2\widehat{\alpha }^2\sigma _4^2(x^*)}{N_k}$, and $u_k:=\frac{(1-3\lambda ^2)(\min \{\lambda \theta ,\widehat{\alpha }\})^2}{2|\mathrm {L}(\xi )|^2_2}\mathfrak {R}^2(F,x^k)$, we have that, almost surely, $\{\Vert x^k-x^*\Vert ^2\}$ converges and $\sum _{k=0}^{\infty }\mathfrak {R}^2(F,x^k)<\infty$. In particular, almost surely, $\{x^k\}$ is bounded and $0=\lim _{k\rightarrow \infty }\mathfrak {R}^2(F,x^k)=\lim _{k\rightarrow \infty }\Vert x^k-\Pi _X[x^k-F(x^k)]\Vert ^2$. This fact and the continuity of F and $\Pi _X$ almost surely imply that every cluster point $\bar{x}$ of $\{x^k\}$ satisfies $\Vert \bar{x}-\Pi _X[\bar{x}-F(\bar{x})]\Vert ^2=0$. Then, we conclude from Lemma 2.1 that $\bar{x}\in X^*$. This together with the boundedness of $\{x^k\}$ and the fact that every cluster point of $\{x^k\}$ belonging to $X^*$ indicate that $\lim _{k\rightarrow \infty }\mathrm{d}(x^k,X^*)=0$ almost surely. In a similar way, we can deduce that $\lim _{k\rightarrow \infty }\mathbb {E}[\mathfrak {R}^2(F,x^k)]=0$ by taking expectation in (4.11). $\square$

Appendix D: Proof of Lemma 4.5

First of all, it follows from Assumption 4.2 that there must exist $k_0$ satisfying (4.12). In the following, we let $d_i:=\Vert x^i-x^*\Vert ^2$ for $i\in \mathbb {N}_0$ and $k\ge k_0$ be given. Taking expectation in (4.11), making use of $\mathbb {E}[\mathbb {E}[\cdot \big |\widehat{\mathcal{F}}_i]]=\mathbb {E}[\cdot ]$, and drop the negative term in the right-hand side. We then sum recursively the obtained inequality from $i:=k_0$ to $i:=k-1$, we obtain

$$\begin{aligned} |d_k|^2_2\le |d_{k_0}|^2_2+\overline{\mathcal{C}}_2\widehat{\alpha }^2\widehat{L}^2_4\sum _{i=k_0}^{k-1}\frac{|d_i|^2_2}{N_i}+\mathcal{C}_2\widehat{\alpha }^2\sigma _4^2(x^*)\sum _{i=k_0}^{k-1}\frac{1}{N_i}.\end{aligned}$$

(A.5)

For any $\varpi >0$, we define the stopping time $t_{\varpi }:=\{k\ge k_0:|d_k|_2>\varpi \}$. From (4.12) and (A.5), we have that, for any $\varpi >0$ such that $t_{\varpi }<\infty$,

$$\begin{aligned}\varpi ^2< & {} |d_{t_{\varpi }}|_2^2\le |d_{k_0}|^2_2+\overline{\mathcal{C}}_2\widehat{\alpha }^2\widehat{L}^2_4\sum _{i=k_0}^{t_{\varpi }-1}\frac{|d_i|^2_2}{N_i}+\mathcal{C}_2\widehat{\alpha }^2\sigma _4^2(x^*)\sum _{i=k_0}^{t_{\varpi }-1}\frac{1}{N_i}\\< & {} |d_{k_0}|^2_2+\phi \varpi ^2+\frac{\phi \mathcal{C}_2\sigma _4^2(x^*)}{\overline{\mathcal{C}}_2\widehat{L}_4^2}.\end{aligned}$$

By the definition of $t_{\varpi }$ for any $\varpi >0$, the argument above implies that any threshold $\varpi ^2$, which $\{\big |d_k\big |^2_2\}_{k\ge k_0}$ eventually exceeds, is bounded above by $\frac{|d_{k_0}|^2_2+\frac{\phi \mathcal{C}_2\sigma _4^2(x^*)}{\overline{\mathcal{C}}_2\widehat{L}_4^2}}{1-\phi }$. As a result, $\{|d_k|^2_2\}_{k\ge k_0}$ is bounded and satisfies the statement of the lemma. $\square$

Appendix E: Proof of Theorem 4.2

Since $\{N_k\}$ satisfies Assumption 4.2, the conclusions in Theorem 4.1 and Lemma 4.5 hold. In particular, $\{x^k\}$ is bounded in $\mathcal{L}^2$. Let M satisfies that $\sup _{k\ge 0}\big |\Vert x^k-x^*\Vert \big |_2^2\le M$. Hence, we have $\sup _k\mathbb {E}[\Vert x^k-x^*\Vert ^2]\le M$. In the recursion of Lemma 4.2, we take expectation, use $\mathbb {E}[\mathbb {E}[\cdot \big |\widehat{\mathcal{F}}_i]]=\mathbb {E}[\cdot ]$, and summing recursively from $i:=0$ to $i:=k$, we have

$$\begin{aligned}&\frac{(1-3\lambda ^2)(\min \{\lambda \theta ,\widehat{\alpha }\})^2 }{2|\mathrm {L}(\xi )|^2_2}\sum _{i=0}^k\mathbb {E}[\mathfrak {R}(F,x^i)^2]\\&\qquad \le \Vert x^0-x^*\Vert ^2+\left( \mathcal{C}_2\widehat{\alpha }^2\sigma _4(x^*)^2+\overline{\mathcal{C}}_2\widehat{\alpha }^2\widehat{L}_4^2M\right) \sum _{i=0}^k\frac{1}{N_i}.\end{aligned}$$

Taking the above inequality, the bound

$$\begin{aligned}\sum _{i=0}^k\frac{1}{N_i}\le \sum _{i=0}^{\infty }\frac{1}{N_i}\le \int ^{\infty }_{-1}\frac{dz}{N(z+\mu )(\ln (z+\mu ))^{1+b}} =\frac{1}{Nb(\ln (\mu -1))^b}\end{aligned}$$

and $\min _{i=0,\cdots ,k}\mathbb {E}[[\mathfrak {R}(F,x^i)^2]\le \frac{1}{k+1}\sum _{i=0}^k\mathbb {E}[[\mathfrak {R}(F,x^i)^2]$, we obtain the conclusion immediately. $\square$

Appendix F: Proof of Theorem 4.3

First, it follows from Theorem 4.2 that there exists a constant $\mathcal{M}>0$ such that, for every $k\in \mathbb {N}$,

$$\begin{aligned} \min _{i=0,\cdots ,K}\mathbb {E}[[\mathfrak {R}(F,x^i)^2]\le \mathcal{M}n(Nbk)^{-1}. \end{aligned}$$

Hence, given $\tau >0$, we obtain $\min _{i=0,\cdots ,K}\mathbb {E}[[\mathfrak {R}(F,x^i)^2]\le \tau$ after $K=\mathcal{O}(nN^{-1}b^{-1}\tau ^{-1})$ iterations. The total number of oracle calls after K iterations is upper bounded by

$$\begin{aligned} \sum _{i=0}^K(1+l_i)N_i\lesssim & {} \left( \max _{i=0,\cdots ,K}l_i\right) \sum _{i=0}^KN_i(\ln i)^{1+b}\lesssim \left( \max _{i=0,\cdots ,K}l_i\right) K^2N(\ln K)^{1+b}\nonumber \\\lesssim & {} \left( \max _{i=0,\cdots ,K}l_i\right) N^{-1}n^2b^{-2}\tau ^{-2}(\ln (nN^{-1}b^{-1}\tau ^{-1}))^{1+b} \end{aligned}$$

(A.6)

and $\min _{i=0,\cdots ,K}\mathbb {E}[[\mathfrak {R}(F,x^i)^2]\le \tau$. By Lemma 4.1, one can obtain that $l_k\le \log _{\frac{1}{\theta }}\left( \frac{\widehat{\alpha }\overline{L}_k}{\min \{\lambda \theta ,\widehat{\alpha }\}}\right)$. This fact, (A.6) and $N=\mathcal{O}(n)$ imply the claimed bound on $\sum _{i=0}^K(1+l_i)N_i$.

The concavity of the mapping $z\mapsto \log _{\frac{1}{\theta }}(z)$ and the Jensen’s inequality imply

$$\begin{aligned} \mathbb {E}[l_i]\le \mathbb {E}\left[ \log _{\frac{1}{\theta }}\left( \frac{\widehat{\alpha }\overline{L}_k}{\min \{\lambda \theta ,\widehat{\alpha }\}}\right) \right] \le \log _{\frac{1}{\theta }}\left( \frac{\widehat{\alpha } L}{\min \{\lambda \theta ,\widehat{\alpha }\}}\right) , \end{aligned}$$

where the last inequality follows from $\mathbb {E}[\overline{L}_k]=L$ by the definitions of $\overline{L}_k$, L and Assumption 4.2. Taking expectation in (A.6), making use of the above relation, and the fact that $N=\mathcal{O}(n)$, implies the claimed bound on $\sum _{i=0}^K(1+\mathbb {E}[l_i])N_i$. $\square$